UVA1345 Jamie's Contact Groups(最大流+二分)

题意:

有n(n ≤ 1000)个人和m(m≤500)个组。一个人可能属于很多组。现在请你从某些组中去掉几个人，使得每个人只属于一个组，并使得人数最多的组中人员数目最小。

分析:看到最多的最小,就可以猜到是一个二分答案ans的题,最大流是没想到...囧.

这题用最大流做,若Ai属于Bi,则连一条Ai到Bi的边,边值为1,然后连超级源点S到Ai的边,边值为1,表示仅有一个Ai,然后对于超级源点T,连Bi到T的边,边值为ans,表示Bi组织最多有ans个人,求最大流,若最大流答案等于n,则表示在每组人限制ans个的情况下能够分配组员到各组,若不等于,则说明在每组人限制ans个的情况下,不能分配组员到各组;

// File Name: 1345.cpp
// Author: Zlbing
// Created Time: 2013/4/22 14:14:53

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)

const int MAXN=2005;
    int n,m;
struct Edge{
    int from,to,cap,flow;
};
bool cmp(const Edge& a,const Edge& b){
    return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct Dinic{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[MAXN];
    bool vis[MAXN];
    int d[MAXN];
    int cur[MAXN];
    void init(int n){
        this->n=n;
        for(int i=0;i<=n;i++)G[i].clear();
        edges.clear();
    }
    void AddEdge(int from,int to,int cap){
        edges.push_back((Edge){from,to,cap,0});
        edges.push_back((Edge){to,from,0,0});//当是无向图时，反向边容量也是cap,有向边时，反向边容量是0
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BFS(){
        CL(vis,0);
        queue<int> Q;
        Q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!Q.empty()){
            int x=Q.front();
            Q.pop();
            for(int i=0;i<G[x].size();i++){
                Edge& e=edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int DFS(int x,int a){
        if(x==t||a==0)return a;
        int flow=0,f;
        for(int& i=cur[x];i<G[x].size();i++){
            Edge& e=edges[G[x][i]];
            if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0)break;
            }
        }
        return flow;
    }
    //当所求流量大于need时就退出，降低时间
    int Maxflow(int s,int t,int need){
        this->s=s;this->t=t;
        int flow=0;
        while(BFS()){
            CL(cur,0);
            flow+=DFS(s,INF);
            if(flow>need)return flow;
        }
        return flow;
    }
    //最小割割边
    vector<int> Mincut(){
        BFS();
        vector<int> ans;
        for(int i=0;i<edges.size();i++){
            Edge& e=edges[i];
            if(vis[e.from]&&!vis[e.to]&&e.cap>0)ans.push_back(i);
        }
        return ans;
    }
    void Reduce(){
        for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
    }
    void ClearFlow(){
        for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
    }
};
vector<int> T[MAXN];
Dinic solver;
bool solve(int mid)
{
    solver.init(n+m);
    int s=0,t=n+m+1;
    REP(i,1,n)
    {
        solver.AddEdge(s,i,1);
        REP(j,0,T[i].size())
        {
            solver.AddEdge(i,n+T[i][j]+1,1);
        }
    }
    REP(i,1,m)
    {
        solver.AddEdge(n+i,t,mid);
    }
    int ans=solver.Maxflow(s,t,INF);
    if(ans==n)return true;
    else return false;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0&&m==0)break;
        REP(i,0,n)T[i].clear();
            char ch[20];
        REP(i,1,n)
        {
            scanf("%s",ch);
            char c;
            int a;
            c=getchar();
                //printf("aaacccc%caaa\n",c);
            while(c!='\n')
            {
                scanf("%d",&a);
                //printf("aa%d\n",a);
                T[i].push_back(a);
                c=getchar();
                //printf("cccc%caa\n",c);
            }
        }
        int L=0,R=n;
        while(L<R)
        {
            int mid=(R+L)/2;
            printf("mid--%d\n",mid);
            if(solve(mid))
            {
                R=mid;
            }
            else{
                L=mid+1;
            }
        }
        printf("%d\n",L);
    }
    return 0;
}